A random variable X has density f(x) = c x(1 − x) for 0 ≤ x ≤ 1 and 0 elsewhere, where c is a constant. Find P(X < 1/3). Give the answer as a fraction in lowest terms.

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Probability from a density

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A random variable $X$ has density $f (x) = c x (1 - x)$ for $0 \leq x \leq 1$ and $0$ elsewhere, where $c$ is a constant. Find $P (X < \frac{1}{3})$ . Give the answer as a fraction in lowest terms.

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Hint

First choose $c$ so that the density integrates to $1$ .

Solution

The integral of $x - x^{2}$ from $0$ to $1$ is $\frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ , so $c = 6$ .
The distribution function is $F (a) = 3 a^{2} - 2 a^{3}$ for $0 \leq a \leq 1$ .
At $a = \frac{1}{3}$ , this is $3 \times \frac{1}{9} - 2 \times \frac{1}{27} = \frac{7}{27}$ .
So $P (X < \frac{1}{3}) = \frac{7}{27}$ .

Answer: $\frac{7}{27}$

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